$\dfrac{ 10a + 7b }{ 9 } = \dfrac{ -a + 9c }{ -3 }$ Solve for $a$.
Multiply both sides by the left denominator. $\dfrac{ 10a + 7b }{ {9} } = \dfrac{ -a + 9c }{ -3 }$ ${9} \cdot \dfrac{ 10a + 7b }{ {9} } = {9} \cdot \dfrac{ -a + 9c }{ -3 }$ $10a + 7b = {9} \cdot \dfrac { -a + 9c }{ -3 }$ Reduce the right side. $10a + 7b = {9} \cdot \dfrac{ -a + 9c }{ -{3} }$ $10a + 7b = -{3} \cdot \left( -a + 9c \right)$ Distribute the right side $10a + 7b = -{3} \cdot \left( -{a} + {9c} \right)$ $10a + 7b = {3}a - {27}c$ Combine $a$ terms on the left. ${10a} + 7b = {3a} - 27c$ ${7a} + 7b = -27c$ Move the $b$ term to the right. $7a + {7b} = -27c$ $7a = -27c - {7b}$ Isolate $a$ by dividing both sides by its coefficient. ${7}a = -27c - 7b$ $a = \dfrac{ -27c - 7b }{ {7} }$